Problem: Solve for $n$. $\left(7^2\right)^{4}=n^8$ $n=$
Answer: The general rule for powers of powers is $\left(x^m\right)^{n}=x^{m\cdot n}$. Let's expand the powers for $ \left({7^2}\right)^{{4}}=n^8}$. $\begin{aligned} \left({7^2}\right)^{4}&=\underbrace{{7^2\cdot 7^2 \cdot7^2 \cdot 7^2}}_{\text{{4 times}}} \\\\\\ &=\underbrace{ \underbrace{{7\cdot 7}}_\text{2 times} \cdot \underbrace{{7\cdot 7}}_\text{2 times} \cdot \underbrace{{7\cdot 7}}_\text{2 times} \cdot \underbrace{{7\cdot 7}}_\text{2 times}} _{\text{4 times}} \\\\ &=\underbrace{n\cdot n\cdot n\cdot n\cdot n\cdot n\cdot n \cdot n }_\text{8 times}} \\\\ \end{aligned}$ $n = 7$